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What minimum radius must a large conducting sphere of an electrostatic generating machine have if it is to be at 20000 V without discharge into the air?

How much charge will it carry?

User SeKa
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1 Answer

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First of all,

The electric field at the surface of the sphere is given by


E = kQ/r²

The field strength at which breakdown occurs in the air is 3.0 MV/m

So, E = 3.0 MV/m

The sphere potential is defined as

V = kQ/r


If we divide E/V we get

E = V/r


r = V/E = 20000V / 3.10^6 V/m = 6.66 10^-3 m = 6.66 mm

2. Charge

V = kQ/r .............>>

Q = Vr/k = 20000V *( 6.66 10^-3 m)/ (9.10^9 N m2/C2) = 1.481 10^-8 C

User Flavius
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